How to Prove a Function is Injective(one-to-one) Using the Definition

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How to prove a function is injective. Injective functions are also called one-to-one functions. This is a short video focusing on the proof.

[@mayankkhare7686]

In just 3 minutes, I gained more knowledge than I did in 4 hours of class. Thanks a lot

[@TheMathSorcerer]

:)

[@og_enthusiast]

bro fr , no hate but that shows your dumb , i mean you could have said 30 minuets , but you are telling me you sat there infront of teacher explaining this for 4 hours and you didnt understand , n***ga what

[@iamplaceholder]

Could use more difficult examples. This example is the one literally everyone covers and doesn't help at all in figuring out difficult functions.

[@TheMathSorcerer]

I have a few more in my playlist I think. Go to my channel and scroll down and check the functions playlist. I think I have a few more. Will make more harder ones😄😄😄😄

[@maxhickman7447]

fkn savage haha

[@daved1113]

These videos on Injunctive and surjunctive proofs were super helpful. Thanks.

[@TheMathSorcerer]

You're very welcome!

[@gautamganesh4794]

Thanks so much for this!! This cleared all my doubts!

[@jingyiwang5113]

Thank you for this amazing video! I was so confused with how to write a clear and precise proof for my homework. Now I am not confused anymore. Thanks!😀

[@TheMathSorcerer]

[@sky_island]

Thankyou! This explanation was excellent

[@TheMathSorcerer]

glad it helped!!

[@jcasma]

Grear video man. Would have loved if you had included some numeric examples, but it's good nonetheless.

[@garbagecuber6502]

Literally so easy to understand! Great video!

[@thexxmaster]

I love you, thanks a bunch

[@TheMathSorcerer]

lol np at all

[@FPrimeHD1618]

Another great video!

[@TheMathSorcerer]

Glad it helped!!

[@ScottRachelson777]

Can you do some proofs involving function spaces and not just individual functions?

[@theonewhoisfluff745]

This ultimately proves that there are no two elements with same function and just 1 element with a unique image right?

[@jonathoncliffbailey]

Hey man thanks a lot! A lot easier to follow than Khan Academy.

[@jonathoncliffbailey]

+Jonathon Bailey Plus the shortness of your explanations means a lot!

[@erniesings6855]

Are all definitions if an only if?

[@TheMathSorcerer]

Yes they are! ALL definitions by default are if and only if.

[@vicente7338]

how would you execute this proof for a piecewise function?

[@TheMathSorcerer]

By taking cases on the elements

[@TheMathSorcerer]

I think I have an example I will look

[@TheMathSorcerer]

Found it kzbin.info/www/bejne/bnaWnnylaKp5q6M

[@vicente7338]

@@TheMathSorcerer yesss thank you i took a look at it and it was clear as day. As an additional question how would I formally find the maximum domain of a function as to make it 1 to 1? I think I have a certain intuition but I cant put my finger on it.

[@nuche3931]

Amazing video How do you prove that f(a+b)=(a+b,a-b) is injective

[@abhishekthakur786]

IF R is a set of real numbers then show that the function f:R-> R defined by f(x) = -sin x, is neither one-one nor onto

[@fernandalara547]

THANK YOU SM :')

[@biswashkhatiwada2293]

At last my confusioon has been cleared.

[@TheMathSorcerer]

:)

[@eriannewyvestarter4972]

But the graph is parabola and it's coming from two different x's. Injective is one-to-one, and the proof isn't satisfying the definition.

[@ivanovmario_5398]

he redefines the domain from all real numbers to all real numbers

[@emigames6248]

Problem: what about using contrapositives?

[@lanl1845]

idk how my prof can make this shit a 4 hour presentation , thank you sir

[@govindjangid6201]

Thanks sir ji

[@TheMathSorcerer]

:)

[@Subscribesful]

what about: g(x) = x^2-6x+10 where g: ]-∞, 3] ↦ [1,∞[ ?

[@truthkmgmailcom]

So quadratics are never injective? (Unless they have a domain restriction)

[@talibanchristian]

TRUE

[@sagnikmazumder3108]

What about y=x/(1+x^2)^(1/2)

[@rick06081]

but if you graph y = x^2, there are more than one x values that correspond with the same y value????

[@eazyvegas]

That is why the domain is restricted to >=0

[@rick06081]

eazyvegas ohh i see, thanks

[@borophyllius]

In this example, the domain is restricted to [0, ∞), so there is only one value. If the domain was not restricted, then f(x) = x^2 is not injective.

[@KingGisInDaHouse]

For those that want an approach that’s a little more obvious. f(x)=x^2. x>=0 in order for this to be on to one f(x+h) must not equal f(x) in any other circumstance other than h=0 or h not being real. f(x+h)= (x+h)^2=x^2+2xh+h^2 =f(x)+2xh+h^2 it’s not the same as f(x) unless 2xh+h^2 is zero 2xh+h^h=0 h(2x+h)=0 h=0 2x+h=0 h=-2x There are points were f(x+h) = f(x). Where h=-2x. f(x-2x)=f(-x) However since we said x>=0 X is positive -1x is just it’s opposite and it’s disqualified so we ignore this case it is one to one.

[@eriktruong9856]

a = b, if a=1 and b=2 would that mean 1=2?

[@inthebackwiththerabbish]

love u

[@TheMathSorcerer]

Thx

[@englishlife5838]

This wrong

[@cheese7119]

Math is hard :'l

[@AB-vi8zl]

Copied Khan academy

[@kootin]

who asked?